Why absence of surjection with the power set is not enough to prove the existence of an undecidable language?

From this statement


As there is no surjection from $\mathbb{N}$ onto $\mathcal{P}(\mathbb{N})$, thus there must exist an undecidable language.

I would like to understand why similar reasoning does not work with a finite set $B$ which also has no surjection onto $\mathcal{P}(B)$! (with $|B|=K$ and $K \in \mathbb{N}$)

Why is there a minimum need for the infinite set?

EDIT Note:

Although I chose an answer, many answers and all comments are important.

Let us first recapitulate in which context the cited statement makes sense.

• Let us restrict ourselves to the domain of (decision) functions in $\mathbb{N} \to \{0,1\}$ (a subset of all functions).

• Every such function corresponds to one element of $\mathcal{P}(\mathbb{N})$ (see characteristic function).

• There are countably many Turing machines (simple encoding).

By 1. and 2., there are uncountably many functions. Therefore, there are functions that have no corresponding Turing machine, that is they are not computable. There are simply too many functions; this is what is meant by "there is no surjection".

Now, there are only countably many finite sets in $\mathcal{P}(\mathbb{N})$(extension of Cantor's pairing function). Therefore, the same contradiction can not be derived when only considering finite sets.

If you add some infinite sets to your base set so it becomes uncountable, there is no reason to believe some of the finite sets were undecidable; you only know that there are some undecidable sets. In fact, all finite sets are decidable, so the culprits are always infinite sets.

StackExchange Computer Science Q#1993, answer score: 6